$\begingroup$A wheel of fortune is divided into 40 sectors, numbered from 1 to 40. Tickets are sold representing each sector. Tickets are \$1 each. All 40 tickets must be sold before the wheel can be spun. Only the winning ticket receives a \$10 prize. Calculate the probability of winning the \$10 prize in one game and again in the next game.
$\endgroup$23 Answers
$\begingroup$Hint: The results in both games are independent of each other. The probability of two independent events occurring is the product of the probabilities of each individual event.
$\endgroup$$\begingroup$It depends on the number of tickets you have. For one ticket the probability of winning the first time would be $1/40$, and for two wins in a row, $(1/40)^2$ by the multiplication rule (note that the events are independent).
$\endgroup$$\begingroup$Since we're only selling 40 total tickets, one for each section, we know that there will be no duplicate tickets sold. In other words, if you buy a ticket for section X (where x is any number from 1-40), you are the only one who holds that ticket.
The probability of section $X$ landing on the first spin is $\frac{1}{40}$.
The probability of section $X$ landing on the second spin is $\frac{1}{40}$.
Note that each results for each spin are independent of each other. No matter what has happened in the past, the probability of a certain section landing on each spin is always going to be $\frac{1}{40}$.
The probability of two independent events happening successively is the product of each event occurring separately. In this case:
$$ P(Winning \ Twice \ in \ a \ Row) = P(X) * P(X) = \frac{1}{40} * \frac{1}{40} = \frac{1}{160} = 0.625 \% $$
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